Simultaneous And Series Disintegration - Practice Questions & MCQ

• Simultaneous decay- 

As we know that due to radioactive disintegration a radio nuclide transforms into its daughter nucleus. Depending on the nuclear structure and its unstability, a parent nucleus may undergo either $\alpha-$ or $\beta-$ emission. Sometimes a parent nucleus may undergo both types of emission imultaneousonly.

If an element decays to different daughter nuclei with different decay constant $\lambda_1, \lambda_2, \lambda_3, \ldots$ etc. for each decay mode, then the effective decay constant of the parent nuclei can be given as

$
\lambda_{e f f}=\lambda_1+\lambda_2+\lambda_3, \ldots
$

Similarly, for a radioactive element with decay constant $\lambda$ which decays by both $\alpha-$ and $\beta-$ decays given that the probability
for an $\alpha$-emission is $\mathrm{P}_1$ and that for $\beta-$ emission is $\mathrm{P}_2$ the decay constant of the element can be split for individual decay modes. Like in this case the decay constants for $\alpha-$ and $\beta-$ decay separately can be given as

$
\begin{aligned}
& \lambda_\alpha=P_1 \lambda \\
& \lambda_\beta=P_2 \lambda
\end{aligned}
$

• Series decay-

Accumulation of Radioactive element in Radioactive series-

A radioactive element decays into its daughter nuclei until a stable element appears. Consider a radioactive series-

$
A_1 \xrightarrow{\lambda_1} A_2 \xrightarrow{\lambda_2} A_3 \xrightarrow{\lambda_3} \ldots
$

A radioactive element $A_1$ disintegrates to form another radioactive element $A_2$ which in turn disintegrates to another element $A_3$ and so on. Such decays are called Series or Successive disintegration.

Here, the rate of disintegration of $A_1=$ Rate of formation of $A_2$

$
\begin{aligned}
& \frac{-d N_{A 1}}{d t}=\frac{d N_{A 2}}{d t}=\lambda_1 N_{A 1} \\
& \frac{-d N_{A 2}}{d t}=\frac{d N_{A 3}}{d t}=\lambda_2 N_{A 2} \\
& \frac{d N_{A 1}}{d t}=-\lambda_1 N_{A 1} \\
& \frac{d N_{A 2}}{d t}=-\lambda_2 N_{A 2}
\end{aligned}
$
Therefore, net formation of $A_2=$ Rate of disintegration of $A_1$ - Rate of disintegration of $A_2$

$
=\lambda_1 N_{A 1}-\lambda_2 N_{A 2}
$

If the rate of disintegration of $A_1$ becomes equal to the Rate of disintegration of $A_{2,}$ then it is called Radioactive equilibrium. So the equation becomes -

$
\Rightarrow \frac{\lambda_1}{\lambda_2}=\frac{N_{A 2}}{N_{A 1}}=\frac{T_{a v g 2}}{T_{a v g 1}}=\frac{\left(T_{\frac{1}{2}}\right)_2}{\left(T_{\frac{1}{2}}\right)_1}
$